Question

The acceleration of moon with respect to earth is $0.0027ms^{-2}$ and the acceleration of an apple falling on earth’s surface is about $10ms^{-2}$.Assume that the radius of the moon is one fourth of the earth’s radius. If the moon is stopped for an instant and then released, it will fall towards the earth. The initial acceleration of the moon towards the earth will be.

Answer 1

We know that acceleration is the rate at which speed changes with respect to time. We also know that the moon and the earth are attracted towards each other by an invisible force also called the gravitational force, which pulls the two to each other with acceleration as discussed below.

Formula used:
$F=\dfrac{GMm}{r^{2}}$
$F=ma$

Complete step-by-step solution:
From the law of gravitational force, we have $F=\dfrac{GMm}{r^{2}}$, where $G$ is acceleration due to gravity, $M$ is the mass of earth, $m$ is the mass of moon and $r$ is the distance between the two.
Also from newton’s law we know that $F=ma$, where $F$ is the force due the mass $m$ under the acceleration $a$
Given that the acceleration of moon with respect to that of earth is $0.0027ms^{-2}$then from gravitational forces between them is, we can say that
$F=\dfrac{GMm}{r^{2}}$
And the acceleration of apple on earth is $g=10ms^{-2}$, $g=\dfrac{F}{m}=10ms^{-2}$,
Then the gravitational forces on earth is $F=\dfrac{GMm}{R^{2}}$, $R$ is the radius of earth, since the earth attracts the moon with the same $g$, we have
$\implies10= \dfrac{F}{m}=\dfrac{GM}{R^{2}}$
Since the radius of the moon is given to be one fourth of the earth’s radius. Then $r=R+\dfrac{1}{4}R=\dfrac{5R}{4}$
Then, we have $a_m$ which is the acceleration experienced by the moon due to earth as
$a_m=\dfrac{F}{m}=\dfrac{GM}{r^2}$
$\implies a_m=\dfrac{GM\times 16}{25R^2}$
$\implies a_m=\dfrac{16\times 10}{25}$
$\therefore a=6.4ms^{-2}$
Clearly, the acceleration with which the earth pulls the moon $g$ is not the same as the acceleration which the moon pulls the earth $a_m$.

Note: To calculate the initial acceleration of the moon towards the earth, we don’t take into the account the given values$0.0027ms^{-2}$. It is given as a distractor, and does not have any significance in this question. Gravitational force is always attractive in nature.